# Thread: Basic topology question

1. ## Basic topology question

This is relatively easy, but it requires an astute observation. You use this quite a bit later in topology so I guess it's a good challenge question.

NOTE: If you have already seen the solution to this, please don't ruin the fun! This is aimed at people who haven't seen a lot of topology before.

Problem: Define $\mathbb{R}^{\infty}$ to be the set of all square-summable real sequences $\{x_n\}$ equipped with the metric $d\left(\{x_n\},\{y_n\}\right)=\sqrt{\sum_{n=1}^{\i nfty}\left|x_n-y_n\right|^2}$. Prove that this metric space is separable.

Note: Separable means it contains a countable dense subset. Also, while my book calls this $\mathbb{R}^{\infty}$ it is most commonly known as $\ell^2$.

2. Originally Posted by Drexel28
This is relatively easy, but it requires an astute observation. You use this quite a bit later in topology so I guess it's a good challenge question.

NOTE: If you have already seen the solution to this, please don't ruin the fun! This is aimed at people who haven't seen a lot of topology before.

Problem: Define $\mathbb{R}^{\infty}$ to be the set of all square-summable real sequences $\{x_n\}$ equipped with the metric $d\left(\{x_n\},\{y_n\}\right)=\sqrt{\sum_{n=1}^{\i nfty}\left|x_n-y_n\right|^2}$. Prove that this metric space is separable.

Note: Separable means it contains a countable dense subset. Also, while my book calls this $\mathbb{R}^{\infty}$ it is most commonly known as $\ell^2$.

Here is the answer. Try to prove it.

Spoiler:

Consider the set $F$ of all eventually zero rational sequences

3. Spoiler:

Proof: Let $F$ be all eventually zero rational sequences

$F=\{(r_1,r_2,\dots,r_n,0,0,\dots)|r_i \in \mathbb{Q},n\in \mathbb{N}\}.$

Clearly, $F \subseteq \ell^2$ and is countable. I claim $\text{cl}(F)=\ell^2$.

Let $y=(y_k)_{k \in \mathbb{N}} \in \ell^2$ and $\varepsilon>0.$ Pick $n_0 \in \mathbb{N}$ such that $\sum_{k=n_0+1}^{\infty}|y_k|^2 =M_{n_0} < \varepsilon^2.$ Since $\mathbb{Q}$ is dense in $\mathbb{R}$, for each $y_k$ (for $1\le k \le n_0$), pick $r_{0k} \in \mathbb{Q}$ such that

$|r_{0k}-y_k| < \sqrt{\frac{1}{n_0}\left(\varepsilon^2-M_{n_0}\right)} \Longrightarrow |r_{0k}-y_k|^2 < \frac{1}{n_0}\left(\varepsilon^2-M_{n_0}\right)$.

So we have

$\sum_{k=1}^{n_0}|r_{0k}-y_k|^2<\varepsilon^2-M_{n_0}$.

If we let $x_0 = (r_{01},r_{02},\dots,r_{0n_0},0,0,\dots) \in F$, then $d(x_0,y)<\varepsilon$. Therefore, $y$ is a limit point of $F$.